Vector Algebra Test

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Vector Algebra Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

If $$\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$$, $$\overrightarrow{b} = \hat{i} + 2\hat{j} - \hat{k}$$, $$\overrightarrow{c} = \hat{i} + \hat{j} - 2\hat{k}$$, then a vector in the plane of $$\hat{b}$$ and $$\hat{c}$$ whose projection on $$\hat{a}$$ is a magnitude of $$\sqrt{\frac{2}{3}}$$ is

A
$$2\hat{i} + 3\hat{j} - 3\hat{k}$$

B
$$2\hat{i} + 3\hat{j} + 3\hat{k}$$

C
$$-2\hat{i} - \hat{j} + 5\hat{k}$$

D
$$2\hat{i} + \hat{j} + 5\hat{k}$$
Question 2
1 / -0

Let $$\vec { p } $$ and $$\vec { q } $$ be the position vectors of the points $$P$$ and $$Q$$ respectively with respect to origin $$O$$. The points $$R$$ and $$S$$ divide $$PQ$$ internally and externally respectively in the ratio $$2:3$$. If $$\overrightarrow { OR } $$ and $$\overrightarrow { OS } $$ are perpendicular, then which one of the following is correct?

A
$$9{ p }^{ 2 }=4{ q }^{ 2 }$$

B
$$4{ p }^{ 2 }=4{ 9q }^{ 2 }$$

C
$$9p=4q$$

D
$$4p=9q$$

Solution

$$\vec{OR}=\dfrac{3p+2q}{3+2}=\dfrac{1}{5}(3p+2q)$$

Given that $$\vec{OR}$$ and $$\vec{OS}$$ are perpendicular.

$$\Rightarrow \vec{OR} \cdot \vec{OS}=0$$

$$\Rightarrow \dfrac{1}{5}(3p+2q) \cdot (3p-2q)=0$$

$$\Rightarrow 9|p^2|-4|q^2|=0$$

$$\Rightarrow 9p^2=4q^2$$
Question 3
1 / -0

A unit vector a makes an angel $$\Pi /4$$ with the z-axis. If a+i+j is a unit vector, then a can be equal to

A
$$\dfrac { i }{ 2 } -\dfrac { j }{ 2 } +\dfrac { k }{ \sqrt { 2 } } $$

B
$$\dfrac { i }{ 2 } -\dfrac { j }{ 2 } -\dfrac { k }{ \sqrt { 2 } } $$

C
$$-\dfrac { i }{ 2 } -\dfrac { j }{ 2 } +\dfrac { k }{ \sqrt { 2 } } $$

D
None

Solution

Equation of z-axis = $$=0i+0j+\hat { k } $$
a is unit vector at angel $$\Pi /4$$
$$\bar { a } .\bar { z } =\left| \bar { a } \right| \left| \bar { z } \right| \cos { \dfrac { \Pi }{ 4 } } $$
Let $$a=xi+y\lambda +2\hat { k } $$
$$z=\cos { \dfrac { \Pi }{ 4 } } \Rightarrow z=\frac { 1 }{ \sqrt { 2 } } $$
$$u=xi+yj+zk\quad =\quad \delta i+yj+\dfrac { 1 }{ \sqrt { 2 } } k$$
$${ x }^{ 2 }+{ y }^{ 2 }=\dfrac { 1 }{ 2 } $$
$$a+i+j$$ is unit vector
$$\left( x+1 \right) i+\left( y+1 \right) j+\dfrac { 1 }{ \sqrt { 2 } } k$$ is unit vector
$${ (x+1 })^{ 2 }+{ (y+1) }^{ 2 }=\dfrac { 1 }{ 2 } $$
$${ (x+1 })^{ 2 }+{ (y+1) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$$
$${ x }^{ 2 }={ (x+1 })^{ 2 }=1\quad \quad -x=x+1\\ a=-\dfrac { 1 }{ 2 } i-\dfrac { 1 }{ 2 } j+\dfrac { 1 }{ \sqrt { 2 } } k\quad \quad x=-\dfrac { 1 }{ 2 }\quad Similarly \quad y=-\dfrac { 1 }{ 2 }$$
Question 4
1 / -0

Let $$\vec{u},\ \vec{v},\ \vec{w}$$ be such that $$\left| \vec { u } \right| =1$$, $$\left| \vec { v } \right| =2$$, $$\left| \vec { w } \right| =3$$. If the projection of $$\vec{v}$$ along $$\vec{u}$$ is equal to projection of $$\vec{w}$$ along $$\vec{u}$$ and $$\vec{v}$$ and $$\vec{w}$$ are perpendicular to each other then $$\left| \bar { u } -\bar { v } +\bar { w } \right|$$ equals-

A
$$2$$

B
$$\sqrt{7}$$

C
$$\sqrt{14}$$

D
$$14$$

Solution
Question 5
1 / -0

If $$ABCDE$$ is a pentagon, then $$\vec {AB}+\vec {AE}+\vec {BC}+\vec {DC}+\vec {ED}+\vec {AC}$$ equals

A
$$3 \vec {AD}$$

B
$$3 \vec {AC}$$

C
$$3 \vec {BE}$$

D
$$3 \vec {CE}$$

Solution

In the pentagon ABCDE, join AC and AD.
By triangle law,
In $$\Delta ABC$$,
$$\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} $$
In $$\Delta ACD$$,
$$\overrightarrow {AD} + \overrightarrow {DC} = \overrightarrow {AC} $$
In $$\Delta AED$$,
$$\overrightarrow {AE} + \overrightarrow {ED} = \overrightarrow {AD} $$
Now,
$$\overrightarrow {AB} + \overrightarrow {AE} + \overrightarrow {BC} + \overrightarrow {DC} + \overrightarrow {ED} + \overrightarrow {AC} = \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) + \left( {\overrightarrow {AE} + \overrightarrow {ED} } \right) + \overrightarrow {DC} + \overrightarrow {AC} $$
$$ = \overrightarrow {AC} + \left( {\overrightarrow {AD} + \overrightarrow {DC} } \right) + \overrightarrow {AC} $$
$$ = \overrightarrow {AC} + \overrightarrow {AC} + \overrightarrow {AC} $$

$$ = 3\overrightarrow {AC} $$
Question 6
1 / -0

Let $$\overline {a}, \overline {b}$$ be two noncollinear vectors. If $$\overline {OA}=(x+4y)\overline {a}+(2x+y+1)\overline {b}, \overline {OB}=(y-2x+2)\overline {a}+(2x-3y-1)\overline {b}$$ and $$3\overline {OA}=2\overline{OB}$$, then $$(x,y)=$$

A
$$(1,2)$$

B
$$(1,-2)$$

C
$$(2,-1)$$

D
$$(-2,-1)$$

Solution
Question 7
1 / -0

The value of $$|\overrightarrow A + \overrightarrow B - \overrightarrow C + \overrightarrow D |$$ can be zero if :-

A
$$|\overrightarrow A | = 5,\,|\overrightarrow B | = 3,|\overrightarrow C | = 4;|\overrightarrow D | = 12$$

B
$$|\overrightarrow A | = 2\sqrt 2 ,\,|\overrightarrow B | = 2,|\overrightarrow C | = 2;|\overrightarrow D | = 5$$

C
$$|\overrightarrow A | = 2\sqrt 2 ,\,|\overrightarrow B | = 2,|\overrightarrow C | = 2;|\overrightarrow D | = 10$$

D
$$|\overrightarrow A | = 5,\,|\overrightarrow B | = 4,|\overrightarrow C | = 3;|\overrightarrow D | = 8$$

Solution
Question 8
1 / -0

The projection of $$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$ on the vector $$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$$ is?

A
$$\sqrt{\dfrac{2}{3}}$$

B
$$\dfrac{2}{\sqrt{21}}$$

C
$$\sqrt{\dfrac{3}{2}}$$

D
$$\dfrac{5}{\sqrt{21}}$$

Solution

The given vectors are $$\vec { a } =\hat { i } +2\hat { j } +3\hat { k }$$ and $$\vec { b } =\hat { i } +2\hat { j } -\hat { k }$$

Let us first find the dot product of the given vectors as shown below:

$$\vec { a } \cdot \vec { b } =(1\times 1)+(2\times 2)+(3\times -1)=1+4-3=5-3=2$$

Now, we find the magnitude of the vector $$\vec { b }$$ as follows:

$$\left| \vec { b } \right| =\sqrt { { 1 }^{ 2 }+2^{ 2 }+(-1)^{ 2 } } =\sqrt { 1+4+1 } =\sqrt { 6 }$$

We know that the projection of $$\vec { a }$$ on $$\vec { b }$$ is $$\dfrac { \vec { a } \cdot \vec { b } }{ \left| \vec { b } \right| }$$, therefore, we have:

$$\dfrac { \vec { a } \cdot \vec { b } }{ \left| \vec { b } \right| } =\dfrac { 2 }{ \sqrt { 6 } } =\dfrac { 2 }{ \sqrt { 2\times 3 } } =\dfrac { 2 }{ \sqrt { 2 } \times \sqrt { 3 } } =\dfrac { \sqrt { 2 } }{ \sqrt { 3 } } =\sqrt { \dfrac { 2 }{ 3 } }$$

Hence, the projection of the vector $$\vec { a }$$ on $$\vec { b }$$ is $$\sqrt { \dfrac { 2 }{ 3 } }$$.
Question 9
1 / -0

The set of values of $$'c'$$ for which the angle between the vectors $$(cx\hat{i}-6\hat{j}+3\hat{k})$$ and $$(x\hat{i}-2\hat{j}+2cx\hat{k})$$ is acute for every $$x\in R$$ is

A
$$(0,\ 4/3]$$

B
$$[0,\ 4/3)$$

C
$$(11/9,\ 4/3)$$

D
$$[0,\ 4/3]$$

Solution
Question 10
1 / -0

A vector $$\vec{a}$$ has components $$2$$ p and $$1$$ with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system, $$\vec{a}$$ components $$p+1$$ and $$1$$, then?

A
$$p=0$$

B
$$p=1$$ or $$p=-\dfrac{1}{3}$$

C
$$p=-1$$ or $$p=\dfrac{1}{3}$$

D
$$p=1$$ or $$p=-1$$

Solution

$$\textbf{Step-1: Find the magnitude of the given vector & simplify.}$$

$$\text{We know that,}$$

$$\text{The magnitude of the vector must remain same irrespective of the co-ordinate system.}$$

$$\implies\quad \left| \overline { a } \right| =\left| { \overline { { a }^{ ' } } } \right|$$

$$ \left| 2p\overset { \wedge }{ i } +\overset { \wedge }{ j } \right| =\left| (p+1)\overset { \wedge }{ { i }^{ ' } } +\overset { \wedge }{ { j }^{ ' } } \right| $$

$$\implies\quad \sqrt { { (2p) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { { (p+1) }^{ 2 }+{ 1 }^{ 2 } } $$

$$\textbf{Step-2: Simplify the above results to get the required unknown.}$$

$$\text{After squaring on both sides we get,}$$

$$\implies\quad { 4p }^{ 2 }+1={ p }^{ 2 }+1+2p+1$$

$$\implies\quad { 4p }^{ 2 }+1={ p }^{ 2 }+2p+2$$

$$\implies\quad { 3 }p^{ 2 }-2p-1=0$$

$$\implies\quad { 3p }^{ 2 }-3p+p-1=0$$

$$\implies\quad 3p(p-1)+1(p-1)=0$$

$$\implies\quad (p-1)(3p+1)=0$$

$$\implies\quad p=1,\cfrac { -1 }{ 3 } $$

$$\textbf{Hence, option - B is the correct answer.}$$